If $n$ is a positive integer such that $n=3j+2$ for some integer $j$, then $n\neq k^2$ for an integer $k$.

Question

So far I have the following:

By Contrapositive: If $n = k^2$ for an integer $k$, then $n \neq 3j+2$ for some integer $j$, if $n$ is a positive integer. We can assume that $n$ is a perfect square for $k$ and $\sqrt{n}$ is an integer. For the sake of contradiction we assume that $n=k^2$ and $n=3j+2$, allowing us to arrive at $k^2=3j+2$. This means there must be an integer $j$ for which $3j+2=k^2$, a perfect square. However if we write the equation as $j=\frac{k^2}{3}-\frac{2}{3}$ there is no integer k that will yield anything other than a fraction for the value of j. This contradicts our assumption that j is an integer proving our fact.

I think my issue is with this statement: $j=\frac{k^2}{3}-\frac{2}{3}$. I am unsure how to prove this will always yield a fraction. I appreciate any help.

Answer 1

Assume that $k^2 = 3j+2$. Obviously $k$ is not divisible by $3$. We have $(k-1)(k+1) = 3j+1$. Now, either $k=3l+1$ or $k=3l+2$. In either case, one of $k-1$ and $k+1$ is divisible by $3$, but $3j+1$ isn't.

Answer 2

$\pmod3$, $k=0,1$ or $2$. $\therefore k^2\cong0,1$ or $4$. So $k^2\cong0$ or $1$.